Max, Min, Sup, Inf - Purdue University

1 CHAPTER 6 Max, Min, Sup, InfWe would like to begin by asking for the maximum of the functionf(x) = (sinx)/x. An approximate graph is indicated below. Lookingat the graph, it is clear thatf(x) 1 for allxin the domain , 1 is the smallest number which is greater than alloff (sin x)/x1 Figure 1 Loosely speaking, one might say that 1 is the maximum value off(x). The problem is that one is not a value off(x) at all. Thereis noxin the domain offsuch thatf(x) = 1. In this situation,we use the word supremum instead of the word maximum . Thedistinction between these two concepts is described in the a set of real numbers. An upper boundforSis a numberBsuch thatx Bfor allx S. The supremum,if it exists, ( sup , LUB, least upper bound ) ofSis the smallest81826. MAX, MIN, SUP, INFupper bound forS.

2 An upper bound which actually belongs to the setis called a that a certain numberMis the LUB of a setSis oftendone in two steps:(1) Prove thatMis an upper bound forS show thatM sfor alls S.(2) Prove thatMis theleastupper bound forS. Often this isdone by assuming that there is an >0 such thatM is also an upper bound forS. One then exhibits an elements Swiths > M , showing thatM is not an the least upper bound for the following set andprove that your answer is {12,23,34, .. ,nn+ 1..}Solution:We note that every element ofSis less than 1 sincenn+ 1<1We claim that the least upper bound is 1. Assume that 1 is notthe least upper bound ,. Then there is an >0 such that 1 is alsoan upper bound . However, we claim that there is a natural numbernsuch that1 <nn+ inequality is equivalent with the following sequence ofinequali-ties1 nn+ 1< 1n+ 1< 1 < n+ 11 1< MAX, MIN, SUP, INF83 Reversing the above sequence of inequalities shows that ifn >1 1,then 1 <nn+1showing that 1 is not an upper bound verifies our a set has a maximum, then the maximum will also be a supre- thatBis an upper bound for a setSand thatB S.

3 ThenB= >0 be given. ThenB cannot be an upper boundforSsinceB SandB > B , showing thatBis indeed theleastupper the least upper bound for the following set andprove that your answer is {1,12,23,34, .. ,nn+ 1..}.Solution:From the work done in Example 1, 1 is an upper boundforS. Since 1 S, 1 = the max, min, sup, and inf of the following setand prove your {2n+ 1n+ 1|n N}.Solution:We write the first few terms ofS:S={32,53,74,95,116, ..}.The smallest term seems to be32and there seems to be no largestterm, although all of the terms seem to be less than 2. Since limn 2n+1n+1=2 we conjecture that:(a) There is no maximum, (b) supS= 2, and (c) minS= infS= MAX, MIN, SUP, INFWe must first show that 2 is an upper bound + 1n+ 1<22n+ 1<2n+ 21<2which is always true. Reversing the above argument shows that 2isan upper we show that 2 is theleastupper bound .

4 If 2 is not theLUB, there is an >0 such 2 is an upper bound . However, weclaim that there aren Nsuch that2 <2n+ 1n+ 1showing that 2 is not an upper prove our claim, note that the above inequality is equivalentwith <2n+ 1n+ 1 2 >1n+ 1n >1 1 Since there existn Nsatisfying the above inequality, our claim isproved. Hence 2 is the we show that 2 is not a maximum. This means showing thatthere is nonsuch that2 =2n+ 1n+ however is equivalent with2(n+ 1) = 2n+ 12n+ 2 = 2n+ 12 = 1which is certainly we prove that minS=32. We first note that32 Ssince32=2 1 + 11 + MAX, MIN, SUP, INF85 Hence, it suffices to show that32is a lower bound which we do asfollows:2n+ 1n+ 1 322(2n+ 1) 3(n+ 1)n 1which is true for alln N. Reversing the above argument shows that32is a lower bound .

5 The central question in this section is Does every non-empty setof numbers have a sup? The simple answer is no the setNof naturalnumbers does not have a sup because it is not bounded from we change the question: Does every set of numbers which isbounded from above have a sup? The answer, it turns out, dependsupon what we mean by the word number . If we mean rationalnumber then our answer is NO!.Recall that the set of integers is the set of positive and naturalnumbers, together with 0. {0, 1, 2, .. , n, |n N}.The set of rational numbers is the setQ={pq|p, q Z, q6= 0}.Thus, for example,23and 97are elements ofQ. In Chapter 9 (The-orem 2) we prove that 2 is not , letSbe the set of all positiverationalnumbersrsuch thatr2<2. Since the square root function is increasing on the set ofpositive real numbers,S={0< r < 2|r Q}.

6 Clearly, 2 is an upper bound forS. It is also a limit of valuesfromS. In fact, we know that 2 = +.Each of the numbers , , , , etc. is rational andhas square less than 2. Their limit is 2. Thus, supS= 2. (SeeExercise 6 below.) Since 2 is irrational,Sis then an example of aset of rational numbers whosesupis , however, that we (like the early Greek mathematicians)only knew about rational numbers. We would be forced to say thatS866. MAX, MIN, SUP, INFhas no sup. The fact thatSdoes not have a sup inQcan be thoughtof as saying that the rational numbers do not completely fill up thenumber line; there is a missing number directly to the right fact that the setRof all real numbersdoesfill up the line is sucha fundamentally important property that we take it as an axiom: thecompleteness axiom.

7 (The reader may recall that in Chapter I, wementioned that we would eventually need to add an axiom to ourlist. This is it.) We shall also refer to this axiom as theLeast UpperBound Axiom.(LUB Axiom for short.)Least upper bound Axiom:Every non-empty set of real numberswhich is bounded from above has a observation that the least upper bound axiom is false forQtells us something important:it is not possible to prove the leastupper bound axiom using only the axioms stated in Chapters 1 is because the set of rational numbers satisfy all the axiomsfrom Chapters 1 and 2. Thus, if the least upper bound axiom wereprovable from these axioms, it hold for the rational course, similar comments apply to minimums:Definition:LetSbe a set of real numbers. A lower bound forSis a numberBsuch thatB xfor allx S.

8 The infinum ( inf , GLB, greatest lower bound ) ofS, if it exists, is the largest lowerbound forS. A lower bound which actually belongs to the set is calleda , once we have the LUB Axiom, we do not need an-other axiom to guarantee the existence of inf s. The existence of inf sis a theorem which we will leave as an exercise. (Of course, we couldhave let the existence of inf s be our completeness axiom, in whichcase the existence of sup s would be a theorem.)Greatest lower bound Property:Every non-empty set of realnumbers which is bounded from below has a that a certain numberMis the GLB of a setSis similarto a LUB proof. It requires:6. MAX, MIN, SUP, INF87(1) Proving thatMis a lower bound forS proving thatM sfor alls S.(2) Proving thatMis thegreatestlower bound forS.

9 Often thisis done by assuming that there is an >0 such thatM+ is a lower bound forS. One then exhibits an elementsofSsatisfyings < M+ , showing thatM+ is not a lowerbound that the inf ofS= (1,5] is :By definitionSis the set ofxsatisfying 1< x 5. Hence1 is a lower bound forS. Suppose that 1 is not the GLB ofS. Thenthere is an >0 such that 1 + is also a lower bound forS. Tocontradict this, we exhibitx Ssuch that 1< x <1 + . Since0< 2< we see thatx= 1 + 2satisfies1< x <1 + .Since 1+ is (by assumption) a lower bound forSand 5 S, 1+ 5,showing thatx (1,5]. Thus, 1 + is not a lower bound , provingthat 1 is the greatest lower upper and lower bounds fory=f(x) forx [ 1, ] wheref(x) = x4+ 2x2+xUse a graphing calculator to estimate the least upper bound and thegreatest lower bound forf(x).))

10 Solution:From the triangle inequality|f(x)|=| x4+ 2x2+x| |x4|+|2x2|+|x|=|x|4+ 2|x|2+|x|The last quantity is largest when|x|is largest, which occurs when|x|= Hence|f(x)| + 2( )3+ = MAX, MIN, SUP, INFH ence,M= 14 is an upper bound andM= 14 is a lower a check, we graphy= x4+2x2+xwith xmin= .5, xmax= ,ymin= 15 and ymax= 15, as well as the linesy= 14 andy= the graph lies between the lines, the value ofMis acceptable,although considerably larger than necessary. To estimate the leastbound, we trace the curve using the trace feature of the calculator,finding that the maximum and minimumy-values are and .130 respectively. These values are (approximately) theleast upper bound and the greatest lower bound :It is important to note that in the preceding example,the values of the function at the end points of the interval are notbounds for the function becausef(x) is not monotonic ( it isneither increasing nor decreasing) over the stated interval.